Warm-up: Divide using Long Division HW: Page 239 (7-12 by any method, 24, 27, 30, 33, 36, 39) page 240 (49, 51, 55, 57, 59)

Homework Answers: Polynomial Long Division 1) 5x - 1 2) 6x + 5 3) 4x2 + 7x + 12 + 4) x2 - 2x - 7 5) 5x2 + 10x + 22 + 1) (15x2 + 22x – 5)  (3x + 5) 2) (12x2 - 32x – 35)  (2x – 7) 3) (4x3 - 2x - x2 + 6)  (x – 2) 4) (3x3 - 5x2 - 23x – 7)  (3x + 1) 5) (5x3 + 2x – 3)  (x – 2)

2.3 Dividing Polynomials Day2 Synthetic Division Objective: Divide a polynomial by a polynomial using synthetic division. Use the remainder theorem to evaluate higher degree polynomials for a given value. Use the factor theorem to factor a polynomial completely.

- 3 1 6 8 -2 - 3 - 9 3 1 x2 + x 3 - 1 1 1 Synthetic Division There is a shortcut for long division as long as the divisor is x – k where k is some number. (Can't have any powers on x). Set divisor = 0 and solve. Put answer here. x + 3 = 0 so x = - 3 1 - 3 1 6 8 -2 Multiply these and put answer above line in next column Multiply these and put answer above line in next column Multiply these and put answer above line in next column Bring first number down below line - 3 Add these up - 9 3 Add these up Add these up 1 x2 + x 3 - 1 1 This is the remainder Put variables back in (one x was divided out in process so first number is one less power than original problem). So the answer is: List all coefficients (numbers in front of x's) and the constant along the top. If a term is missing, put in a 0.

Let's try another Synthetic Division 0 x3 0 x Set divisor = 0 and solve. Put answer here. x - 4 = 0 so x = 4 1 4 1 0 - 4 0 6 Multiply these and put answer above line in next column Multiply these and put answer above line in next column Multiply these and put answer above line in next column Multiply these and put answer above line in next column Bring first number down below line 4 Add these up 16 48 192 Add these up Add these up Add these up 1 x3 + x2 + x + 4 12 48 198 This is the remainder Now put variables back in (remember one x was divided out in process so first number is one less power than original problem so x3). So the answer is: List all coefficients (numbers in front of x's) and the constant along the top. Don't forget the 0's for missing terms.

Let's try a problem where we factor the polynomial completely given one of its factors. You want to divide the factor into the polynomial so set divisor = 0 and solve for first number. - 2 4 8 -25 -50 Multiply these and put answer above line in next column Multiply these and put answer above line in next column Multiply these and put answer above line in next column Bring first number down below line - 8 Add these up 50 Add these up Add these up No remainder so x + 2 IS a factor because it divided in evenly 4 x2 + x - 25 Put variables back in (one x was divided out in process so first number is one less power than original problem). So the answer is the divisor times the quotient: List all coefficients (numbers in front of x's) and the constant along the top. If a term is missing, put in a 0. Check to see if it factors further

Find f(3) for the following polynomial function. f(x) = 5x2 – 4x + 3 f(3) = 5(3)2 – 4(3) + 3 f(3) = 5 ∙ 9 – 12 + 3 f(3) = 45 – 12 + 3 f(3) = 36

Now divide the same polynomial by (x – 3). 5x2 – 4x + 3 3 5 –4 3 15 33 5 11 36

The Remainder Theorem f(x) = 5x2 – 4x + 3 5x2 – 4x + 3 3 5 –4 3 15 33 5 11 36 f(x) = 5x2 – 4x + 3 f(3) = 5(3)2 – 4(3) + 3 f(3) = 5 ∙ 9 – 12 + 3 f(3) = 45 – 12 + 3 f(3) = 36 Notice that the value obtained when evaluating the function at f(3) and the value of the remainder when dividing the polynomial by x – 3 are the same. 5x2 – 4x + 3 = (5x2 + 11x) ∙ (x – 3) + 36 f(x) = q(x) ∙ (x – a) + f(a)

The Remainder Theorem If a polynomial f(x) is divided by (x – a), the remainder is the constant f(a), and f(x) = q(x) ∙ (x – a) + f(a) where q(x) is a polynomial with degree one less than the degree of f(x).

The Remainder Theorem Use synthetic substitution to find g(4) for the following function. f(x) = 5x4 – 13x3 – 14x2 – 47x + 1 4 5 –13 –14 –47 1 20 28 56 36 5 7 14 9 37

The Factor Theorem The binomial (x – a) is a factor of the polynomial f(x) if and only if f(a) = 0.

The Factor Theorem When a polynomial is divided by one of its binomial factors, the quotient is called a depressed polynomial. If the remainder of the depressed polynomial is zero, that means f(#) = 0. This means that the divisor resulting in a remainder of zero is a factor of the polynomial.

The Factor Theorem x3 + 4x2 – 15x – 18 x – 3 3 1 4 –15 –18 3 21 18 3 1 4 –15 –18 3 21 18 1 7 6 0 Since the remainder is zero, (x – 3) is a factor of x3 + 4x2 – 15x – 18. This also allows us to find the remaining factors of the polynomial by factoring the depressed polynomial.

The Factor Theorem x3 + 4x2 – 15x – 18 x – 3 3 1 4 –15 –18 3 21 18 3 1 4 –15 –18 3 21 18 1 7 6 0 The factors of x3 + 4x2 – 15x – 18 are (x – 3)(x + 6)(x + 1). x2 + 7x + 6 (x + 6)(x + 1)

The Factor Theorem (x – 3)(x + 6)(x + 1). Compare the factors of the polynomials to the zeros as seen on the graph of x3 + 4x2 – 15x – 18.

The Factor Theorem Use synthetic division to show that x is a solution of the third-degree polynomial equation. Use the result to factor the polynomial completely. List all real solutions of the equation. x3 – 11x2 + 14x + 80 = 0 x = 8 (x – 8)(x – 5)(x + 2) x = 8, 5, -2

Sneedlegrit: Use synthetic division to show that x is a solution of the third-degree polynomial equation. Use the result to factor the polynomial completely. List all real solutions of the equation. 2. 2x3 + 7x2 – 33x – 18 = 0 x = – 6 (x + 6)(2x + 1)(x – 3) x = -6, -1/2, 3 HW: Page 239 (7-12 by any method, 24, 27, 30, 33, 36, 39) page 240 (49, 51, 55, 57, 59)